Question

Consider the concentration cell in which the metal ion has a charge of +2, and the solution concentrations are: dilute solution = 0.05 M concentrated solution = 4.808 M What is the predicted Ecell, using the Nernst equation? ____ V

Answer 1

Answer: The predicted cell potential of the cell is +0.0587 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  $M(s)\rightarrow M^{2+}+2e^-$

Reduction half reaction (cathode):  $M^{2+}+2e^-\rightarrow M(s)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[M^{2+}_{(diluted)}]}{[M^{2+}_{(concentrated)}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

$[M^{2+}_{(diluted)}]$ = 0.05 M

$[Zn^{2+}_{(concentrated)}]$ = 4.808 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{2}\log \frac{0.05M}{4.808M}$

$E_{cell}=0.0587V$

Hence, the predicted cell potential of the cell is +0.0587 V

Answer 2

Using Nernst equation, the predicted cell voltage is 0.59 V.

From Nernst equation;

Ecell = E° - 0.592/n log Q

In this particular case;

Ecell = E° - 0.592/n log [dilute]/[concentrated]

Also, since the metal is has a charge of +2;

M^2+(aq) + 2e -----> M(s)

Now from the question;

[dilute] = 0.05 M

[concentrated] = 4.808 M

n = 2

E° = 0 V

Substituting values;

Ecell = 0 - 0.592/2 log [0.05 M]/[4.808 M]

Ecell = 0.59 V

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